Proposition 22. Theorem.

564. Two opposite polyedral angles are symmetrical.






Hyp. Let S - ABC, S-A'B'C' be two opp. triedral _.8.
To prove they are symmetrical.
Proof. Because the face ZS ASB and A'SB' are vertical

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... ZASB = LA'SB'.

(53) Similarly, BSC = _B'SC', etc.

Also, because the diedral _ between two planes is the same at every pt.,

(528) .:. the diedral Zs whose edges are SA, SB, etc., = respectively the diedral Z8 whose edges are SA', SB', etc.

But the edges of S- A'B'C' are arranged in the reverse order from the edges of S-ABC. :.S - ABC is symmetrical to S - A'B'C'.

Q. E.D.


Pass two parallel planes, one through each of two straight lines which do not meet and are not parallel.

Let AB, CD be the lines : draw AE ll to CD, CF || to AB. . . plane A EB is || to plane CFD.

Proposition 23. Theorem. 565. The sum of any two face-angles of a triedral angle is greater than the third.


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To prove

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B Hyp. Let S - ABC be a triedral / in which ¿ASC is the greatest face L.

ZASB + BSC > LASC. Proof. In the plane ASC draw SD, making

_ ASD = LASB. Draw AC cutting SD in D, take SB = SD, and join AB, BC.

Because the As ASB, ASD have SA common, SB = SD, LASB = LASD, (Cons.) 5. AASB=A ASD,.. AB = AD.

(104) In Δ ABC, AB + BC > AC.

(96) But

AB = AD. (Proved above) Subtracting, BC > DC.

(Ax. 5) Because in the As BSC, DSC, SC is common, SB = SD, and BC > DC, .. BSC > _DSC.

(120) But LASB = LASD.

(Cons.) Adding, LASB + ZBSC > LASC.


Proposition 24. Theorem.

566. The sum of the face-angles of any convex polyedral angle is less than four right angles.

Hyp. Let the convex polyedral ZS

S be cut by a plane making the section ABCDE a convex polygon.

To prove _ ASB + BSC, etc., < 4 rt. Zs.

Proof. From any pt. 0 within the polygon ABCDE draw 0A, OB, OC, E OD, OE.

C There are thus formed two sets of As, one with their common vertex at S, and the other with their common vertex at 0, and an equal number of each.

.:. the sum of the Zs of these two sets of As is equal. (97)

Because the sum of any two face Zs of a triedral Z > the third,

(565) ... ZSAE + SAB > ZEAB,


ZSBA + ZSBC > ZABC, etc.

Taking the sum of these inequalities, we find that the sum of the Zs at the bases of the As whose common vertex is S > the sum of the Zs at the bases of the As whose common vertex is 0.

... the sum of the Zs at S < the sum of the Zs at 0.

... the sum of the Zs at S < 4 rt. Zs.

(56) Q.E.D. Proposition 25. Theorem. 567. Two triedral angles, which have the three faceangles of the one equal respectively to the three face-angles of the other, are either equal or symmetrical. S



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Hyp. In the triedral /s, S and S’, let the face Z8 ASB,
BSC, CSA = the face Zs A'S'B', B'S'C', C'S'A'.

To prove S - ABC = or symmetrical to S’ - A'B'C'.

Proof. In the edges SA, S'A', etc., take the six equal distances SA, SB, SC, S'A', S'B', S'C'; and join AB, BC, CA, A'B', B'C', C'A'.

. Then As SAB, SBC, SCA = As S'A'B', S'B'C', S'C'A'. .. A ABC = A A'B'C'.

(108) In the edges SA, S'A', take SD = S'D'.

In the faces ASB, ASC, and A’S'B', A'S'C', draw DH, DK, and D'H', D'K' I to AS and A'S', meeting AB, AC and A'B', A'C', in H, K and H', K'. Join HK and H'K'. Because AD = A'D', and ZDAH = _D'A'H', '

' ... rt. A ADH = rt. A A'D'H'.

(107) .:. AH = A'H', and DH = D'H'. In the same way, AK = A'K', and DK = D'K'. .:: AAHK = AA'H'K', and HK = H'K'. .


'. (104) ... = .:: AHDK = AH'D'K', and HDK = _ H'D'K'. (108) ... diedral ZSA = diedral ZS'A'.

(536) Similarly, diedral Zs SB, SC = diedral Zs S'B', S'C'.

Now if the equal s are arranged in the same order, as in the first two figures, the two triedral s are equal. (557)

But if the equal Zs are in reverse order, as in the first and third figures, the triedral Zs are symmetrical. (562)


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THEOREMS. 1. If a straight line is parallel to a plane, any plane perpendicular to the line is perpendicular to the plane. See (503) and (540).

2. If a line is parallel to each of two intersecting planes, it is parallel to their intersection.

3. If a line is perpendicular to a plane, every plane parallel to that line is also perpendicular to the plane.

4. If a line is parallel to each of two planes, the intersections which any plane passing through it makes with the planes are parallel.

5. If a straight line and a plane are perpendicular to the same straight line, they are parallel.

6. If two straight lines are parallel, they are parallel to the common intersection of any two planes passing through them.

7. If the intersections of several planes are parallel, the normals drawn to them from any point are in one plane.

8. If a plane is passed through one of the diagonals of a parallelogram, the perpendiculars to it from the extremities of the other diagonal are equal.

9. Prove that the sides of an isosceles triangle are equally inclined to any plane through the base.

10. From a point P, PA is drawn perpendicular to a given plane, and from A, AB is drawn perpendicular to a line in that plane: prove that PB is also perpendicular to that line.

11. If the projections of any line upon two intersecting planes are each of them straight lines, prove that the line itself is a straight line. See (195).

12. If a plane is passed through the middle point of the common perpendicular to two straight lines in space, and parallel to both of these lines, prove that the plane bisects

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